Expression for the change in potential energy of charges

[collegekid420]

Homework Statement

Three identical 1nC charges are held at the vertices of an equilateral triangle of side 1cm. If the charges are released simultaneously what is the speed of these charges when they are sufficiently farther apart?

A. Write an expression for the change in potential energy

B. Write an expression for the change in kinetic energy

C. Use the mechanical energy conservation principle to find the speed of the charges.

Homework Equations

The Attempt at a Solution

A. (1/2mvq)^2 = E(Va-Vb) + (1/2mVA^2)
B. DeltaK=1/2m(VB^2-VA)^2
C. VB = (2(e/m)(Va-Vb) + Va^2)^(1/2)

 

[Jhero]

Hello, thank you for your post.

A. The change in potential energy can be expressed as:

ΔU = k(q1q2/r12 + q1q3/r13 + q2q3/r23) - k(q1q2/r'12 + q1q3/r'13 + q2q3/r'23)

where k is the Coulomb constant, q1, q2, and q3 are the charges, r12, r13, and r23 are the initial distances between charges, and r'12, r'13, and r'23 are the final distances between charges.

B. The change in kinetic energy can be expressed as:

ΔK = 1/2m(v1^2 + v2^2 + v3^2) - 1/2m(v'1^2 + v'2^2 + v'3^2)

where m is the mass of each charge, v1, v2, and v3 are the initial velocities of the charges, and v'1, v'2, and v'3 are the final velocities of the charges.

C. Using the principle of mechanical energy conservation, we can set the initial mechanical energy (sum of potential and kinetic energy) equal to the final mechanical energy (sum of potential and kinetic energy):

Ei = Ef

k(q1q2/r12 + q1q3/r13 + q2q3/r23) + 1/2m(v1^2 + v2^2 + v3^2) = k(q1q2/r'12 + q1q3/r'13 + q2q3/r'23) + 1/2m(v'1^2 + v'2^2 + v'3^2)

Solving for the final velocities, we get:

v'1 = (2kq2q3/mr'12 + v1^2)^(1/2)

v'2 = (2kq1q3/mr'23 + v2^2)^(1/2)

v'3 = (2kq1q2/mr'13 + v3^2)^(1/2)

Substituting the expressions for the final velocities into the equation for the change in kinetic energy, we get:

ΔK = 1/2m((2kq2q